Community

1. Analyze ImproveDefine Measure Control
LEAN
SIX SIGMA
Chi-Square
Analysis
Chi-Square Training for Attribute Data

2. Bonacorsi Consulting 2
Lean
Six Sigma
What Is A Chi-Square Test?
What is a Chi-Square?
The probability density curve of a chi-square distribution is
asymmetric curve stretching over the positive side of the line
and having a long right tail.
The form of the curve depends on the value of the degrees of
freedom.

3. Bonacorsi Consulting 3
Lean
Six Sigma
Types of Chi-Square Tests?
Types of Chi-Square Analysis:
Chi-square Test for Association is a (non-parametric, therefore can be used
for nominal data) test of statistical significance widely used bivariate tabular
association analysis.
 Typically, the hypothesis is whether or not two different populations are
different enough in some characteristic or aspect of their behavior
based on two random samples.
 This test procedure is also known as the Pearson chi-square test.
Chi-square Goodness-of-fit Test is used to test if an observed distribution
conforms to any particular distribution. Calculation of this goodness of fit test
is by comparison of observed data with data expected based on the particular
distribution.

4. Bonacorsi Consulting 4
Lean
Six Sigma
The Basics
When to apply a Chi-Squared Test:
 Chi-Squared test is used to determine if there is a statistically
significant difference in the proportions for different groups. To
accomplish this, it breaks all outcomes into groups.
What the Chi-Squared Test does:
 It starts by determining how many defects, for example, would be
“expected” in each group involved.
 It does this by assuming that all groups have the same defect rate
(which Minitab approximates from the data provided).
 Minitab then compares the expected counts with what was actually
observed.
 If the numbers are different by a large enough amount, Chi-Square
determines that the groups do not have the same proportion.

5. Bonacorsi Consulting 5
Lean
Six Sigma
The Basics (Cont.)
Chi-Square Requirements:
 Data is typically attribute (discrete). At the very least, all data must
be able to be categorized as being in some category or another).
 Expected cell counts should not be low (definitely not less than 1
and preferable not less than 5) as this could lead to a false positive
indication that there is a difference when, in fact, none exists.
Chi-Square Hypotheses:
 Ho: The null hypotheses (P-Value > 0.05) means the populations
have the same proportions.
 Ha: The alternate hypotheses (P-Value <= 0.05) means the
populations do NOT have the same proportions.

6. Bonacorsi Consulting 6
Lean
Six Sigma
Using Minitab – Step 1
Enter
Category
Names
Enter the Defect
Qty for the Data
Collection Period
Enter the Success
Qty for the Data
Collection Period
Step 1: Enter
data into Minitab

7. Bonacorsi Consulting 7
Lean
Six Sigma
Using Minitab – Step 2
Step 2: Choose Stat, then
Tables, then Chi-Squared
Test (Two-Way Table in
Worksheet…)

8. Bonacorsi Consulting 8
Lean
Six Sigma
Using Minitab – Step 3
Step 3: Enter the subgroup data to compare
Determine the
subgroups you want
to compare for
variation.
In this example we
examine the
variation between
success (C2) &
Defect (C3) for each
Engineering
Change Proposal
(ECP)
Enter category
names you want
to compare
Click help for
Minitab help on
Chi-Square
Choose Ok
to run Chi-
Squared Test
Enter category
names you want
to compare

9. Bonacorsi Consulting 9
Lean
Six Sigma
The Session Window Results
Observed
counts
Expected
counts
Sum of each
category (Machine)
Sum of all
observed
counts
Degree of
Freedom
(df = n-1)
The Chi-Square test
calculates using the
formula
(40 - 45.87)2
/
45.87 = 0.752
Since the P-Value is <= 0.05, there is a
95% chance that there IS a statistically
significant difference, in other words the
groups (Machines) involved do NOT have
the same defect rate.
Machine #5

10. Bonacorsi Consulting 10
Lean
Six Sigma
Chi-Square Statistic
The series of numbers are
added together for the overall
chi-square statistic. When the
value of the Chi-square statistic
is high enough, the differences
between the groups are
concluded to be too large to be
mere chance and the groups are
determined to be truly different.
The series of numbers are
added together for the overall
chi-square statistic. When the
value of the Chi-square statistic
is high enough, the differences
between the groups are
concluded to be too large to be
mere chance and the groups are
determined to be truly different.

11. Bonacorsi Consulting 11
Lean
Six Sigma
Errors in the Chi-Square
Note: if the expected cell counts are below 5, Minitab will print a warning.
The warning is generated because of the fact that with the expected count
in the denominator, a small value potentially creates an artificially large chi-
square statistic. This is particularly troublesome if more than 20% of the
cells have expected counts less than 5 and the contribution to the overall
chi-square statistic is considerable.
Additionally, if any of the expected cell counts are below 1, Minitab will not
even produce a p-value since the chi-square statistic is sure to be artificially
inflated. In either of these cases, the binomial distribution (Minitab: Stat/
ANOVA/ Analysis of Means) may be able to be used.
Lastly: Attribute Gage R&R (AR&R) or Kappa Test is needed with an
acceptable level of measurement system error prior to running a Chi-Square
Analysis

12. Bonacorsi Consulting 12
Lean
Six Sigma
Lets See How You Do
Example test questions for Analyze
1. Consider a certain operation which produced 100 documents on line 1, 200
documents on line 2, and 300 documents on line 3. Additionally, it is known
that group 1 produced 17 defective documents, group 2 produced 8 defective
documents, and group 3 produced 11 defective documents. If a chi-square
test were done on this data, what would be the expected number of defects
seen by group 3 if the null hypothesis were completely true?
a) 24
b) 18
c) 12
d) 6
2. Using the same data set, is this data statistically significant? Yes or No
Enter data in to Minitab like this:
Group Defective Acceptable
(group 1) 17 83
(group 2) 8 192
(group 3) 11 289
Note: Answers are located in the note section of this
slide

13. Bonacorsi Consulting 13
Lean
Six Sigma
Tips and Tricks
Tips:
 Determine the subgroups and categories to be tested for variation
(differences in proportions) as part of your data collection plan.
 Define the operational definitions for success/defect, the stratifications layers
(subgroups) and the Cause & Effect diagram (fishbone) to pre-determine where
the team believes differences in proportions may exist.
 Continuous (Variable) data can usually be converted into Discrete (Attribute) data
by using categories (Example: cycle time (continuous 1 hr, 1.5 hr, 2 hr) can be
categorized into Cycle Time Met = 1 where success is cycle time <= 8 hrs or
Cycle Time Not Met = 0 where a defect is cycle time > 8 hrs.)
Tricks
 An (MSA) Attribute R&R (Kappa Analysis) for discrete data or Gage R&R for
continuous (variable) data is used prior to calculating the Chi-Square Test to
ensure that the measurement variation < 10% Contribution. If the measurement
variation is > 10% then the variation you will see in the Chi-Square Test is not
valid as too much of the variation seen is coming from your measurement system
(10% MSA error) and not your process variation.

14. Analyze ImproveDefine Measure Control
LEAN
SIX SIGMA
Chi-Square
Analysis
Appendix A
Backup Slides

15. Bonacorsi Consulting 15
Lean
Six Sigma
Chi-Square (X2) Distribution
Row for 95% confidence
Degree of Freedom
(df) = n-1
Machine example
for Success or
Defect per Month
where we had n =
10 (Machines)
Df = 10 – 1 = 9.
Machine Example Summary
Chi-Sq = 0.752 + 0.637 +
0.814 + 0.689 +
2.393 + 2.028 +
1.391 + 1.179 +
1.471 + 1.247 +
0.255 + 0.217 +
0.376 + 0.319 +
1.492 + 1.264 +
0.312 + 0.264 +
0.752 + 0.637 = 18.489
DF = 9, P-Value = 0.030
The sum of the
Chi-Squared Test
(18.489 would
need to be > 16.92
where df = 9
To be 95% confident
of a difference, the
P-Value <= 0.05
Chi Squared Distribution table is used by Minitab to calculate the P-Value using the Degree of
Freedom (df) and the variation between each category’s data sets being compared for differences.
The sum of the
Chi-Squared Test
(18.489 would need
to be > 16.92 where
df = 9) for Ho
Degree of Freedom
(df) = n-1

16. Bonacorsi Consulting 16
Lean
Six Sigma
Chi-Square (for Two-Way)
* For expected frequencies less than 5 the calculated chi-square value changes dramatically
as fe changes. Therefore the calculated value becomes less reliable and should be
interpreted cautiously.
Ho: Independent (There is no relationship between the populations)
Ha: Dependent (There is a relationship between the populations)
Rejection Criteria:
Fail to reject Ho when p ≥ 0.05; Accept Ha when p < 0.05
Compare a calculated value to a critical values from a χ2
table.
Rejection Criteria:
Fail to reject Ho when p ≥ 0.05; Accept Ha when p < 0.05
Compare a calculated value to a critical values from a χ2
table.
Chi-Square equation for two-way
tables:
Where:
f0 = observed frequency
fe = expected frequency
r = number of rows
c = number of columns
g = number of groups = r * c
degrees of freedom = (r-1) * (c-1)
All fe’s should be > 5 *
e
χ =
(f
e
j = 1
g 2
f
o f )
Σ2

17. Bonacorsi Consulting 17
Lean
Six Sigma
Bonacorsi Consulting
Steven Bonacorsi, President
Lean Six Sigma Master Black Belt
14 Clinton Street
Salem NH 03079
sbonacorsi@comcast.net
603-401-7047